原文链接: https://leetcode-cn.com/problems/merge-in-between-linked-lists

英文原文

You are given two linked lists: list1 and list2 of sizes n and m respectively.

Remove list1's nodes from the ath node to the bth node, and put list2 in their place.

The blue edges and nodes in the following figure indicate the result:

Build the result list and return its head.

 

Example 1:

Input: list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
Output: [0,1,2,1000000,1000001,1000002,5]
Explanation: We remove the nodes 3 and 4 and put the entire list2 in their place. The blue edges and nodes in the above figure indicate the result.

Example 2:

Input: list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
Output: [0,1,1000000,1000001,1000002,1000003,1000004,6]
Explanation: The blue edges and nodes in the above figure indicate the result.

 

Constraints:

• 3 <= list1.length <= 104
• 1 <= a <= b < list1.length - 1
• 1 <= list2.length <= 104

中文题目

给你两个链表 list1 和 list2 ，它们包含的元素分别为 n 个和 m 个。

请你将 list1 中下标从 a 到 b 的全部节点都删除，并将list2 接在被删除节点的位置。

下图中蓝色边和节点展示了操作后的结果：

请你返回结果链表的头指针。

 

示例 1：

输入：list1 = [0,1,2,3,4,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
输出：[0,1,2,1000000,1000001,1000002,5]
解释：我们删除 list1 中下标为 3 和 4 的两个节点，并将 list2 接在该位置。上图中蓝色的边和节点为答案链表。

示例 2：

输入：list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
输出：[0,1,1000000,1000001,1000002,1000003,1000004,6]
解释：上图中蓝色的边和节点为答案链表。

 

提示：

• 3 <= list1.length <= 104
• 1 <= a <= b < list1.length - 1
• 1 <= list2.length <= 104

通过代码

高赞题解

解题思路

此处撰写解题思路

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
           ListNode l1=list1;
            ListNode l2=list2;
            ListNode tmp=new ListNode(0);
            int i=0,j=0,z=0;
            while(l1.next != null){
                if(i == a-1){
                    tmp=l1.next;
                    l1.next=l2;
                    break;
                }else{
                    l1=l1.next;
                    i++;
                }

            }

        while(tmp != null){
            if(i+j == b){

                break;
            }else{
                tmp=tmp.next;
                j++;
            }
        }
        ListNode l11=list1;
        while(l11 != null){
                l11=l11.next;
                if(l11.next == null){
                    l11.next=tmp;
                    break;
                }
        }
        return list1;
    }
}

统计信息

通过次数	提交次数	AC比率
13806	18240	75.7%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言